# Show bijectivity of simple map

I know that the following can be done easier using different argument but I'd like to just proof it the "hard way" for learning experience.

$f_a: G\to G, \ \ x\mapsto a\star x,\quad a\in G$

Injectivity:

Claim: $a\neq b \Rightarrow f(a)=f(b)$

Proof: Choose $a,b,c \in G$ s.t. $b\neq c$ now: $f_a(b)=a\star b$ and $f_a(c)=a \star c$

since $b\neq c \Rightarrow f_a(b)=f_a(c)$

So the function is injective.

Surjectivity:

Claim: $\forall b \in G \ \exists c \in G : f(c)=b$

Proof: Choose $a, b, c \in G$ with with $b=a\star c$

Now: $b=a\star c = f_a(c)$

Which fullfilsl the claim, thus the function is surjective.

Since it is injective and surjective, it is also bijective.

Does that hold like that?

• $G$ is a group? Jan 19 '18 at 9:10
• What is $f$ here? Is it the function that sends $a$ to function $f_a$ maybe? Further if you want to prove injectivity of some function $g$ then you must prove that $a\neq b\implies g(a)\neq g(b)$. So I really do not understand your claim that says $a\neq b\implies f(a)=f(b)$. Jan 19 '18 at 9:10

What you did was wrong. There is also context missing, so I suggest you edit your post. I assume that $G$ is a group.

To show injectivity of a function $f: X \to Y$, you have to show:

$$\forall x,y \in X: (x \neq y \implies f(x) \neq f(y))$$

or equivalently (by the principle of contraposition):

$$\forall x,y \in X: (f(x) = f(y) \implies x =y)$$

So assume $f_a(x) = f_a(y)$ for $x,y \in G$. Then by definition of $f_a$ it follows that $a \star x = a \star y$ and by multiplying both sides (on the left) with the inverse $a^{-1}$ (which exists because $G$ is a group), it follows, using associativity and the definition of neutral element, that $x = y$. Hence, the function is injective.

In your proof of surjectivity, you say "choose $a,b,c \in G$ such that $b = a \star c$. You can't assume this! This is exactly what you have to prove! How do you know such elements exist?

So you should rather do the following:

Let $y \in G$. Let $x = a^{-1} \star y \in G$. Then $f_a(x) = f_a(a^{-1} \star y) = a\star a^{-1} \star y = y$

Again, I used associativity and the definition of neutral element implicitely. It's up to you to fill in the details.

So we have proven: $\forall y \in G: \exists x \in G: f_a(x) = y$, which means that $f_a$ is surjective.

Alternative method: You can notice that the mapping $g: G \to G: x \mapsto a^{-1}x$ is a well defined function and that it is the inverse of the given mapping. Hence, the function you are given is bijective (as it is invertible).

I don't want to sound harsh here, but you should maybe take a step back from group theory for some time, and revise some more basics about functions. Make a lot of exercises on injectivity and surjectivity, because these are things that pop up all the time in every field of mathematics.

As there is a comment below your post which states that there is a little bit of confuison of what map you are considering, I will say few words about it before answering your question.

So you consider an action of a group $G$ on itself. That is, you take an element $a$ of a group, $a \in G$, and you consider the mapping $$f_a \colon G \rightarrow G$$ given by multiplication on the right by this element, i.e. for arbitrary $g \in G$ we have $$f_a (g) = a \star g \ ,$$ where $\star$ is the binary operation of your group. It is not hard to show that this is a group homorphism and it follows directly from the axioms of a group. In a similar fashion we can show that this map is actually a bijection (and thus a group isomorphism).

Let us recall the definition of a bijection and injection for an arbitrary map $g \colon A \rightarrow B$ between sets Definition of a bijection: A map $g$ is said to be a bijection if it is both surjective and injective.
Definition of an injection: A map $g$ is said to be an injection if the following implication holds for arbirary elements $b,c$ of the domain of $g$ $$b \neq c \Rightarrow g(b) \neq g(c)$$ So in this definition you had a flow since you considered equality on the right side of the implication sign (such an implication, if satisfied, would violate the injectivity of the map)

The definition of a surjection you wrote is correct thus I will not write it again.

Now we proceed with checking your proof. 1) injectivity: this is wrong due to incorrect definition. let me show how this can be done considering proof by contrapositive, i.e. proving the following implication, equivalent to our original implication above (for more about proof by contrapositive see this wiki article) $$f_a (b) = f_a(c) \Rightarrow b = c \ .$$ Using the definition of $f_a$ we can write $$f_a (b) = a \star b \text{ and }\ \ f_a(c) = a \star c \ ,$$ thus we have $$a \star b = a \star c \ .$$ Multiplying by $a^{-1}$ (the inverse of $a$ which we know exists due to the fact we work in a group) from the left we get $$a^{-1} \star a \star b = a^{-1} \star a \star c$$ (recall that the group operation is associative so bracketing of terms is unnecessary). The above is equivalent to the desired equivality $$b = c$$ (as $a^{-1} \star a = e = a \star a^{-1}$ where $e$ is the neutral element of the group) and thus $f_a$ is injective.

2) surjectivity: You assumed that the element $b$ can be written as a product $a \star c$ for some $c$. This is not incorrect but may be considered as incomplete for you should show that such an element $c$ always exists. Once you do this you are OK. So how to find this $c$? (recall that $a$ is fixed an $b$ is arbitrary). You can always choose $c = a^{-1} \star b$ since then $a \star c = a \star a^{-1} \star b = b$. Then of course for each $b$ this element $c$ is the preimige with respect to $f_a$ since (as you wrote) $f_a (c) = a \star a^{-1} \star b = b$ and thus $f$ is surjective.